Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 640-660, October 2010
ELA
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ON THE GAPS IN THE SET OF EXPONENTS OF BOOLEAN
PRIMITIVE CIRCULANT MATRICES∗
MARIBEL I. BUENO† AND SUSANA FURTADO‡
Abstract. In this paper, we consider the problem of describing the possible exponents of boolean primitive
circulant matrices. We give a conjecture for the possible such exponents and prove this conjecture in several cases.
In particular, we consider in greater detail the case of matrices whose generating vector has three nonzero entries.
Key words. Circulant primitive matrices, Exponent of primitive matrices, Circulant digraphs, Basis for a group.
AMS subject classifications. 11P70, 05C25, 05C50.
1. Introduction. A Boolean matrix is a matrix over the binary Boolean algebra {0, 1}.
An n-by-n Boolean matrix C is said to be circulant if each row of C (except the first one) is
obtained from the preceding row by shifting the elements cyclically 1 column to the right. In
other words, the entries of a circulant matrix C = (cij ) are related in the manner: ci+1,j =
ci,j−1 , where 0 ≤ i ≤ n − 2, 0 ≤ j ≤ n − 1, and the subscripts are computed modulo n.
The first row of C is called the generating vector. Here and throughout, we number the rows
and columns of an n-by-n matrix from 0 to n − 1.
The set of all n-by-n Boolean circulant matrices forms a multiplicative commutative
semigroup Cn with |Cn | = 2n [3, 8]. In 1974, K.H. Kim-Buttler and J.R. Krabill [6], and S.
Schwarz [9] investigated this semigroup thoroughly.
An n-by-n Boolean matrix C is said to be primitive if there exists a positive integer k
such that C k = Jn , where Jn is the n-by-n matrix whose entries are all ones and the product
is computed in the algebra {0, 1}. The smallest such k is called the exponent of C, and we
denote it by exp(C). Let us also denote by En the set {exp(C) : C ∈ Cn , C is primitive}.
In [1], we stated the following question: Given a positive integer n, what is the set En ?
The previous question can easily be restated in terms of circulant graphs or bases for
finite cyclic groups, as we show next.
∗ Received by the editors December 7, 2009. Accepted for publication August 22, 2010. Handling Editor: Bryan
L. Shader.
† Department of Mathematics, University of California, Santa Barbara, CA, USA (mbueno@math.ucsb.edu).
Supported by Dirección General de Investigación (Ministerio de Ciencia y Tecnología) of Spain under grant
MTM2009-09281.
‡ Faculdade de Economia do Porto, Rua Dr. Roberto Frias, 4200-464 Porto, Portugal (sbf@fep.up.pt).
640
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 640-660, October 2010
ELA
http://math.technion.ac.il/iic/ela
641
On the Exponent of Boolean Primitive Circulant Matrices
Let C be a Boolean primitive circulant matrix, and let S be the set of positions corresponding to the nonzero entries in the generating vector of C, where the columns are
counted starting with zero instead of one. C is the adjacency matrix of the circulant digraph
Cay(Zn , S). The vertex set of this graph is Zn and there is an arc from u to u + a (mod n)
for every u ∈ Zn and every a ∈ S. A digraph D is called primitive if there exists a positive
integer k such that for each ordered pair a, b of vertices, there is a directed walk from a to b
of length k in D. The smallest such integer k is called the exponent of the primitive digraph
D. Thus, a circulant digraph is primitive if and only if its adjacency matrix is. Moreover, if
they are primitive, they have the same exponent. Therefore, finding the set En is equivalent
to finding the possible exponents of circulant digraphs of order n.
Let S be a nonempty subset of the additive group Zn . For a positive integer k, we denote
by kS the set given by
kS = {s1 + s2 + · · · + sk : si ∈ S} ⊂ Zn .
The set kS is called the k-fold sumset of S.
The set S is said to be a basis for Zn if there exists a positive integer k such that kS = Zn .
The smallest such k is called the order of S, denoted by order(S). It is well known that the
set S = {s0 , s1 , . . . , sr } ⊂ Zn is a basis if and only if gcd(s1 − s0 , . . . , sr − s0 , n) = 1. We
denote by Sn the set of all bases for Zn .
In [1], we proved that, given a matrix C in Cn , if S is the set of positions corresponding
to the nonzero entries in the generating vector of C, then C is primitive if and only if S is a
basis for Zn . Moreover, if C is primitive, then exp(C) = order(S). Therefore, finding the set
En is equivalent to finding the possible orders of bases for the cyclic group Zn . This question
is quite interesting by itself.
Note that the only primitive matrix in C2 is the 2-by-2 matrix with all entries equal to
1, so E2 = {1}. From now on, we assume that n ≥ 3. In [1], we presented a conjecture
concerning the possible exponents attained by n-by-n Boolean primitive circulant matrices
which we consider here in greater detail.
Given a positive integer n ≥ 3, let c be the smallest positive integer such that
jnk
c
<
¹
º
n
+ c.
c+1
We call c the critical point of n and we denote it by cn . Clearly, cn ≤
(1.1)
¥n¦
2
+ 1.
C ONJECTURE 1. If C is an n-by-n Boolean primitive circulant matrix, then either
exp(C) =
¹ º
n
+ k,
j
(1.2)
Electronic Journal of Linear Algebra ISSN 1081-3810
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642
M.I. Bueno and S. Furtado
for some j ∈ {1, 2, . . . , cn − 1} and k ∈ {−1, 0, 1, . . . , j − 2}, or
¹ º
n
+ cn − 2.
exp(C) ≤
cn
(1.3)
Moreover, for every m ≤ ⌊n/cn ⌋ + cn − 2, there exists a matrix whose exponent is m.
In the literature, the problem of computing all possible exponents attained by circulant
primitive matrices or, equivalently, by circulant digraphs, has been considered. In [2] and
[11], it is shown that if a circulant matrix C is primitive, then its exponent is either n − 1,
⌊n/2⌋, ⌊n/2⌋ − 1 or does not exceed ⌊n/3⌋ + 1. Matrices with exponents n − 1, ⌊n/2⌋,
⌊n/2⌋ − 1 are also characterized. All these results can be immediately translated into results
about the possible orders of bases for a finite cyclic group. In a recent preprint [5], the authors
prove that if S is a basis for Zn of order greater than k for some positive integer k, then there
exists dk such that the order of S is within dk of n/l for some integer l ∈ [1, k]. Notice that
the result we present in Conjecture 1 produces gaps in the set of orders which are larger than
the ones encountered in [5]. Moreover, we show that our gaps should be maximal. In [5],
the authors also prove the existence of the additional gap [⌊n/4⌋ + 3, ⌊n/3⌋ − 2] although
they do not use the techniques presented in the same paper. See [4] for a detailed proof of the
existence of such gap.
In this paper, we give partial results related with Conjecture 1 and give a class of matrices
for which it is shown that the conjecture holds. All the results in the paper are given in terms
of bases for Zn since the equivalent formulation of the problem in these terms resulted more
fruitful than the original statement of the problem in terms of matrices.
In Section 2, we give the explicit value of the critical point cn , as well as some of its
interesting properties. In Section 3, we define Maximal Generalized Gaps and show that the
set of gaps that follow from Conjecture 1 are maximal. In Section 4, we introduce some
concepts and give some results concerning the order of general bases for Zn . In Section 5,
we give some results about the order of bases for Zn with cardinality 3. These results will
allow us to prove Conjecture 1 for some classes of bases for Zn in Section 6. In Section 7,
we extend some of the results given in Sections 5 and 6 to general bases for Zn . Finally, in
Section 8, we present the conclusions as well as some open questions.
√
√
2. The critical point. In this section, we prove that cn is either ⌊ 3 n⌋ or ⌊ 3 n⌋ + 1.
√
We first show that cn ≤ ⌊ 3 n⌋ + 1.
√
L EMMA 2.1. Let n be a positive integer and r = ⌊ 3 n⌋ . Then
º ¹
º
¹
n
n
<
+ r + 1.
(2.1)
r+1
r+2
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
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On the Exponent of Boolean Primitive Circulant Matrices
643
Proof. Suppose that
¹
º ¹
º
n
n
≥
+ r + 1.
r+1
r+2
(2.2)
We have
n = (r + 2)m + t,
where m =
j
n
r+2
k
and 0 ≤ t < r + 2. Then (2.2) is equivalent to
¹
º
t+m
≥r+1
r+1
which implies that
(r + 1)2 − t ≤ m,
or, equivalently, multiplying by r + 2 and adding t,
(r + 1)2 (r + 2) − t(r + 2) + t ≤ n.
Let us show that
(r + 1)3 ≤ (r + 1)2 (r + 2) − t(r + 2) + t
(2.3)
which leads to a contradiction, as n < (r + 1)3 . Notice that
(r + 1)2 − t(r + 1) ≥ (r + 1)2 − (r + 1)(r + 1) = 0,
which implies (2.3).
√
Next we show that cn ≥ ⌊ 3 n⌋.
√
L EMMA 2.2. Let n be a positive integer and r = ⌊ 3 n⌋ . Let p be an integer such that
0 < p < r. Then
º
¹ º ¹
n
n
>
+ p.
(2.4)
p
p+1
Proof. Suppose that
¹
º
¹ º
n
n
+p≥
.
p+1
p
Then
n≥
¶
µ¹ º
n
− p (p + 1).
p
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 640-660, October 2010
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644
M.I. Bueno and S. Furtado
We will show that
We have
³j k
n
p
´
− p (p + 1) > n, which gives a contradiction. Thus, (2.4) holds.
¶
¹ º
¹ º
µ¹ º
n
n
n
− p (p + 1) =
p+
− p2 − p
p
p
p
¹ º
n
>n−p+
− p2 − p
p
¹ º
n
− p2 − 2p > n,
=n+
p
where the last inequality follows because
j k
n
p
− p2 − 2p > 0, as
n ≥ (p + 1)3 = p3 + 3p2 + 3p + 1 > p3 + 2p2 + p.
It follows from Lemmas 2.1 and 2.2 that the smallest cn such that
√
√
is either ⌊ 3 n⌋ or ⌊ 3 n⌋ + 1.
√
T HEOREM 2.3. Let n be a positive integer and r = ⌊ 3 n⌋ . If
jnk
r
j
n
cn
k
º
n
+r
<
r+1
¹
<
j
n
cn +1
k
+ cn
(2.5)
√
√
then cn = ⌊ 3 n⌋ , otherwise cn = ⌊ 3 n⌋ + 1.
We now give some properties of cn that will be useful later.
L EMMA 2.4. Let n be a positive integer. Then n ≤ (cn + 1)2 (cn − 1).
Proof. If n = (cn + 1)2 (cn − 1) + k for some positive integer k, then
¹
º
¹
º
¹
º ¹ º
n
k
k−1
n
2
2
+ cn = cn + cn − 1 +
≤ cn + cn − 1 +
=
,
cn + 1
cn + 1
cn
cn
which is a contradiction by the definition of cn .
Before presenting the next results we introduce the following notation: If a and b are
integers, with b ≥ a, then [a, b] denotes the set of integers in the real interval [a, b]. Moreover,
[a] denotes the set containing just the integer a. If b < a, then [a, b] = ∅.
The next lemma shows that, if cn ≥ 3, the interval [⌊n/cn ⌋ + 2, ⌊n/(cn − 1)⌋ − 1] is
nonempty if and only if n = 14 or n ≥ 16.
L EMMA 2.5. Let n be a positive integer such that cn ≥ 3. Then ⌊n/(cn − 1)⌋ ≥
⌊n/cn ⌋ + 3 if and only if n = 14 or n ≥ 16.
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
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On the Exponent of Boolean Primitive Circulant Matrices
Proof. If cn = 3 or cn = 4, then the result can be verified by a direct computation.
Suppose that cn ≥ 5. Let k − (cn − 1)3 . By Theorem 2.3, k ≥ 0. Note that
º ¹ º
¹
n
n
≥
+3
(2.6)
cn − 1
cn
if and only if
¹
Since
j
k
cn −1
k
−
j
k−1
cn
k
º ¹
º
k
k−1
≥
+ 5 − cn .
cn − 1
cn
≥ 0, if cn ≥ 5, then the result holds.
The next lemma gives an upper bound for the length of the interval
L EMMA 2.6. Let n be a positive integer such that cn ≥ 3.
√
• If cn = ⌊ 3 n⌋, then
º ¹ º
¹
n
n
−
≤ cn + 3.
cn − 1
cn
√
• If cn = ⌊ 3 n⌋ + 1 and n = c3n − 1, then
º ¹ º
¹
n
n
−
= cn + 2.
cn − 1
cn
√
• If cn = ⌊ 3 n⌋ + 1 and n ≤ c3n − 2, then
º ¹ º
¹
n
n
−
≤ cn + 1.
cn − 1
cn
hj
n
cn
k j
ki
, cnn−1 .
Proof. Let n = pcn + q, where p = ⌊n/cn ⌋ and 0 ≤ q < cn . Notice that
º ¹ º ¹
º
¹
n
p+q
n
−
.
=
cn − 1
cn
cn − 1
√
Suppose that cn = ⌊ 3 n⌋. By Lemma 2.4, n ≤ (cn + 1)2 (cn − 1), and therefore,
¹ º
n
≤ c2n + cn − 2.
cn
Hence,
º
n
+ cn − 1 ≤ c2n + 2cn − 3 = (cn − 1)(cn + 3),
p+q ≤
cn
¹
which implies that
¹
º
p+q
≤ cn + 3.
cn − 1
(2.7)
Electronic Journal of Linear Algebra ISSN 1081-3810
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M.I. Bueno and S. Furtado
√
Suppose that cn = ⌊ 3 n⌋ + 1. Then (cn − 1)3 ≤ n ≤ c3n − 1. If n = c3n − 1,
¹
n
p=
cn
º
= c2n − 1,
q = cn − 1
and
p + q = c2n + cn − 2 = (cn − 1)(cn + 2).
Then
¹
º
p+q
= cn + 2.
cn − 1
√
If cn = ⌊ 3 n⌋ + 1 and c3n − cn ≤ n < c3n − 1, then p = c2n − 1 and q ≤ cn − 2. Therefore,
p + q ≤ c2n − 1 + cn − 2,
which implies
¹
º
p+q
≤ cn + 1.
cn − 1
√
If cn = ⌊ 3 n⌋ + 1 and n ≤ c3n − cn − 1, then
p + q ≤ c2n − 2 + cn − 1,
which implies
¹
º
p+q
≤ cn + 1.
cn − 1
L EMMA 2.7. Let n be a positive integer such that cn ≥ 3. Then
√
• if cn = ⌊ 3 n⌋, then 4cn ≤ ⌊n/cn ⌋ + 3;
√
• if cn = ⌊ 3 n⌋ + 1 and n = c3n − 1, then 3cn ≤ ⌊n/cn ⌋ + 2.
√
Proof. If cn = ⌊ 3 n⌋, then n ≥ c3n , which implies that
¹
n
cn
º
≥ c2n ≥ 4cn − 3,
√
for cn ≥ 3. If cn = ⌊ 3 n⌋ + 1 and n = c3n − 1, then
¹
for cn ≥ 3.
n
cn
º
= c2n − 1 ≥ 3cn − 2,
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On the Exponent of Boolean Primitive Circulant Matrices
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3. Maximal generalized gaps. Let n be a positive integer. Let En = {order(S) : S ∈
Sn }. It is well known [2] that En ⊂ [1, n − 1]. We call a gap in En a nonempty interval
A ⊂ [1, n − 1] such that A ∩ En = ∅. We say that a gap A in En is maximal if A′ ∩ En 6= ∅
for any interval A′ ⊂ [1, n − 1], with A strictly contained in A′ .
For each positive integer n and each j ∈ {1, 2, . . . , cn − 1}, if
¹
º
¹ º
n
n
+j ≤
− 2,
j+1
j
let
Bj,n =
·¹
º
¹ º
¸
n
n
+ j,
−2 ,
j+1
j
(3.1)
otherwise let Bj,n = ∅.
Clearly, if Conjecture 1 is true and Bj,n is nonempty, Bj,n is a gap in En . Though the
intervals Bj,n are not necessarily maximal gaps in En , the next theorem shows that, for each
positive integer j, there is an integer n, with j ≤ cn −1, such that Bj,n is a maximal gap in En .
¥ ¦
Here, we use the result that, if b > 1 is a divisor of n, then order({0, 1, b}) = nb + b − 2,
which is a particular case of Corollary 5.4. If a ∈ Zn , we denote by hai the cyclic group
generated by a in Zn .
T HEOREM 3.1. For each positive integer j, there is an integer n, with j ≤ cn − 1, such
that Bj,n is a maximal gap in En .
Proof. We show that for each j there isj an integer
n, with j ≤ cn − 1, and
k
j ktwo bases for
n
Zn , say S1 and S2 , such that order(S1 ) = j+1
+ j − 1 and order(S2 ) = nj − 1.
Let n = j(j + 1)(j + 3). First, we show that cn − 1 ≥ j. If j = 1, then n = 8 and
√
cn = 3. If j > 1, then n − (j + 1)3 = j 2 − 1 > 0, which implies that 3 n > j + 1. Then
√
√
cn − 1 ≥ ⌊ 3 n⌋ − 1 ≥ 3 n − 2 > j − 1, where the first inequality follows from Theorem 2.3.
k
j
n
+j−1.
Since j+1 divides n, by Corollary 5.4, for S1 = {0, 1, j+1}, order(S1 ) = j+1
If j > 1, let S2 = h(j + 1)(j + 3)i ∪ (1 + h(j + 1)(j + 3)i). Then, for k > 0,
kS2 =
k
[
i=1
(i + h(j + 1)(j + 3)i) .
If
j jk= 1, let S2 = {0, 1}. In any case, it is easy to see that order(S2 ) = (j + 1)(j + 3) − 1 =
n
j − 1.
4. Order of bases for Zn . Let T be a subset of the additive group Zn , and let q ∈ Zn .
We define q + T = {q + t : t ∈ T } and q ∗ T = {qt : t ∈ T }.
Electronic Journal of Linear Algebra ISSN 1081-3810
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M.I. Bueno and S. Furtado
Clearly, if S ⊂ Zn and q ∈ Zn , S is a basis for Zn if and only if q + S is a basis for Zn .
Moreover, if S is a basis, order(S) = order(q + S).
L EMMA 4.1. Let n and q be positive integers. If S ∈ Sn and gcd(q, n) = 1, then
q ∗ S ∈ Sn and order(S) = order(q ∗ S).
Proof. It is enough to show that, for all k ≥ 1, |q ∗ kS| = |kS|, as k(q ∗ S) = q ∗ (kS).
Let T = kS. Clearly, |q ∗ T | ≤ |T |. Now suppose that t1 , t2 ∈ T with t1 6= t2 . Suppose
that qt1 = qt2 (mod n). Then (t1 − t2 )q = 0 (mod n), or equivalently, (t1 − t2 )q = kn for
some positive integer k. Since gcd(q, n) = 1, t1 − t2 = 0 (mod n). As 0 ≤ t1 , t2 < n, then
t1 − t2 = 0, which is a contradiction. Thus, |q ∗ T | ≥ |T |, which completes the proof.
We note that, if gcd(n, q) 6= 1, then q ∗ S is not a basis for Zn .
Let S1 , S2 ⊂ Zn . We say that S1 and S2 are equivalent, and we write S1 ∼ S2 , if there
exist integers q1 and q2 , where gcd(q1 , n) = 1, such that S2 = q2 + q1 ∗ S1 . Note that ∼ is
an equivalence relation. Clearly, from the observations above, if S1 ∈ Sn and S1 ∼ S2 , then
S2 ∈ Sn and order(S1 ) = order(S2 ).
Note that if S = {s1 , s2 , . . . , st } ∈ Sn , then S ∼ {0, s2 − s1 , . . . , st − s1 }. Therefore,
in what follows we assume that 0 ∈ S.
R EMARK 1. Let S = {0, a} ∈ Sn . Then order(S) = n − 1 since S ∼ {0, 1}, as a is a
unit for Zn .
We now introduce some definitions that will be used in the next sections.
Let S = {0, s1 , . . . , st } ∈ Sn . Then any element q ∈ Zn can be expressed as x1 s1 +
· · ·+xt st (mod n), for some nonnegative integers x1 , . . . , xt . Moreover, if q 6= 0, the smallest
k such that q ∈ kS is the minimum x1 +· · ·+xt among all the solutions (x1 , . . . , xt ), xi ≥ 0,
to x1 s1 + · · · + xt st = q (mod n).
D EFINITION 4.2. Let S = {0, s1 , . . . , st } ∈ Sn and q ∈ Zn . If q = 0, then we define
exp(q; S, n) = 1; otherwise we define
exp(q; S, n) := min{x1 + · · · + xt : x1 s1 + · · · + xt st = q (mod n), xi ≥ 0}.
Clearly, if S is a basis for Zn and 0 ∈ S, order(S) = max{exp(q; S, n) : q ∈ Zn }.
D EFINITION 4.3. Let S = {0, s1 , . . . , st } ∈ Sn , q ∈ Zn and k be a positive integer. If
q 6= 0, we say that q is (k; S, n)-periodic if k is the smallest nonnegative integer for which
there are nonnegative integers x1 , . . . , xt satisfying
x1 + · · · + xt = exp(q; S, n) and x1 s1 + · · · + xt st = q + kn.
If q = 0, we say that q is (0; S, n)-periodic. We say that S is K-periodic if there exist
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On the Exponent of Boolean Primitive Circulant Matrices
649
(K; S, n)-periodic elements in Zn and there are no (k; S, n)-periodic elements in Zn for
k > K.
R EMARK 2. If the minimum nonzero element of a basis S of Zn , say b, is not 1, then S
is not 0-periodic as any b′ , with 0 < b′ < b, is not (0; S, n)-periodic.
We finish this section with the following lemma.
L EMMA 4.4. [2] Let S ∈ Sn and m be a divisor of n. Suppose that S contains an
element of order m. Then
order(S) ≤
n
+ m − 2.
m
By Remark 1, all bases for Zn , n ≥ 3, with cardinality 2 have order n − 1, and therefore,
they satisfy Conjecture 1. In the next section we focus on bases with cardinality 3.
5. Order of bases for Zn with cardinality 3. By Sn,r we denote the set of all bases for
Zn with cardinality r.
For a given positive integer n, we define pn as follows:
½
⌊n/2⌋ + 1, if n is odd
pn =
⌊n/2⌋,
if n is even.
(5.1)
L EMMA 5.1. Let S = {0, s1 , s2 } ∈ Sn,3 . If gcd(s1 , n) = 1 or gcd(s2 , n) = 1 or
gcd(s2 − s1 , n) = 1, then there exists b ∈ Zn such that b ≤ pn and S ∼ {0, 1, b}.
Proof. Without loss of generality, suppose that either gcd(s1 , n) = 1 or gcd(s2 −
s1 , n) = 1. In the first case, s1 is a unit in Zn and
−1
S ∼ S1 = s−1
1 S = {0, 1, s1 s2 }.
−1
−1
If s−1
1 s2 ≤ ⌊n/2⌋, the claim holds with b = s1 s2 . If s1 s2 > ⌊n/2⌋, then
S ∼ S2 = 1 − S1 = {0, 1, n + 1 − s−1
1 s2 }
and the claim holds with b + 1 − s−1
1 s2 . In the second case, that is, gcd(s2 − s1 , n) = 1, let
S1 = −s1 + S = {0, s2 − s1 , n − s1 }.
Then S ∼ S2 = s′ S1 , where s′ = (s2 − s1 )−1 . Now the argument used above applies to
show the result.
We note that if gcd(s1 , n) 6= 1, gcd(s2 , n) 6= 1 and gcd(s2 − s1 , n) 6= 1 and one of s1 ,
s2 , s2 − s1 , n − s1 , n − s2 , n − s2 + s1 is a product of a divisor of n and a unit in Zn , then
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there exist a, b ∈ Zn such that a is a divisor of n, a 6= 1 and S = {0, s1 , s2 } ∼ {0, a, b}. To
see this, assume, without loss of generality, that s1 < s2 . Note that
S ∼ S1 = −s1 + S = {0, n − s1 , s2 − s1 }
∼ S2 = −s2 + S = {0, n + s1 − s2 , n − s2 }.
Suppose that s1 is a product of a divisor of n and a unit. The proof is analogous in the other
mentioned cases, eventually by considering S1 or S2 instead of S. If s1 |n, then the result is
clear; otherwise s1 = d1 t1 , where d1 |n and gcd(n, t1 ) = 1. Then
−1
S ∼ t−1
1 S = {0, d1 , t1 s2 },
and the result follows.
We were not able to prove that every basis S ∈ Sn,3 that is not equivalent to a basis
of the form {0, 1, b} is equivalent to a basis {0, a, b} with a 6= 1 a divisor of n. However,
numerical experiments show that if these bases exist, they are rare.
T HEOREM 5.2. Let S = {0, a, b} ∈ Sn,3 , where a is a divisor of n and a 6= 1. Then
a − 1 ≤ order(S) ≤
n
+ a − 2.
a
Proof. The inequality on the right follows from Lemma 4.4.
Consider the quotient map f : Zn → Za . Notice that T = f (S) = {0, f (b)}. Since S is
a basis and a divides n, gcd(a, b) = 1 and f (b) 6= 0. Therefore, T is a basis for Za . Moreover,
by Remark 1, order(T ) = a − 1. Since order(S) ≥ order(T ), we get order(S) ≥ a − 1.
>From now on, we consider bases S of Sn,3 of the form {0, 1, b}. For convenience, we
write exp(q; b, n) instead of exp(q; S, n); we also say that S is 0-periodic instead of (0; S, n)periodic.
If S = {0, 1, b} ∈ Sn,3 , then, considering the standard addition and multiplication in Z,
for k ≥ 1,
kS = [0, k] ∪ [b, b + k − 1] ∪ [2b, 2b + k − 2] ∪ · · · ∪ [(k − 1)b, (k − 1)b + 1] ∪ [kb].
For j = 0, 1, . . . , k, let
Ij,k = [jb, jb + k − j].
T HEOREM 5.3. Let S = {0, 1, b} ∈ Sn,3 . Then
jnk
jnk
≤ order(S) ≤
+ b − 2.
b
b
(5.2)
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¥ ¦
Proof. Let n = mb + t, with m = nb and 0 ≤ t < b. Considering the standard
addition and multiplication in Z, the largest element in kS is kb. Thus, if k = order(S), then
kb ≥ n − 1 = mb + t − 1, which implies the inequality on the left.
We have I0,b−1 ∪ I1,b−1 = [0, 2b − 2] ⊂ (b − 1)S. Moreover, {0, b, . . . , (m − 1)b} ⊂
(m − 1)S. Thus,
((b − 1) + (m − 1))S = (b − 1)S + (m − 1)S = Zn .
which implies the inequality on the right.
We now focus on 0-periodic bases of Sn,3 .
C OROLLARY 5.4. Let S = {0, 1, b} ∈ Sn,3 . If S is 0-periodic, then
jnk
+ b − 2.
order(S) =
b
¥ ¦
Proof. Suppose kS = Zn . Let j0 = nb − 1. Note that k ≥ j0 . Since S is 0-periodic,
necessarily [j0 b, j0 b + b − 1] ⊂ Ij0 ,k , which implies that
³j n k
´
³j n k
´ jnk
−1 b+k−
−1 ≥
b − 1,
b
b
b
¥ ¦
¥ ¦
that is, k ≥ nb + b − 2. Then order(S) ≥ nb + b − 2. Since, by Theorem 5.3, order(S) ≤
¥n¦
b + b − 2, the result follows.
We next characterize the 0-periodic bases in Sn,3 . Note that, by Remark 2, we may
assume that these bases are of the form {0, 1, b}. First, we add a technical lemma.
It is clear that if b and w are positive integers, with b ≥ 2, then the minimum x + y
¥ ¦
among all the solutions of x + by = w, with x, y ≥ 0, is obtained when y is y0 = wb . Since
¥ ¦
x = w − by, the minimum value of x + y is x0 + y0 = w − wb (b − 1). Note also that
x0 = w − by0 < b. We then have the following lemma.
L EMMA 5.5. Let b, q ∈ Zn , with b ≥ 2 and q 6= 0. Then
º
¾
½
¹
q + kn
, k≥0 .
exp(q; b, n) = min q + kn − (b − 1)
b
Note that
exp(q; b, n) ≤ q − (b − 1)
jqk
b
≤ q,
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where the first inequality follows from Lemma 5.5.
L EMMA 5.6. Let S = {0, 1, b} ∈ Sn,3 . Then S is 0-periodic if and only if either b
divides n or
´
³j n k
+ 1 ≤ n.
(5.3)
(b − 1)
b
¥ ¦
Proof. Suppose that S is 0-periodic and b is not a divisor of n. Let q = nb b − 1. By
Lemma 5.5 and Definition 4.3,
¹
º
jqk
n+q
≤ q + n − (b − 1)
.
(5.4)
q − (b − 1)
b
b
¥
¦
¥ ¦
¥ ¦
= 2 nb . Also,
As n + q = 2 nb b + (t − 1), for some 1 ≤ t < b, it follows that n+q
b
¥q¦ ¥n¦
b = b − 1. Thus, (5.4) is equivalent to (5.3).
To prove the converse note that, from Lemma 5.5, if b divides n, S is 0-periodic, as, for
any k > 0,
kn
> 0.
b
Now suppose that (5.3) holds. According to Lemma 5.5, we need to show that, for any k > 0
and any q ∈ Zn \{0},
º
¹
jqk
q + kn
,
q − (b − 1)
≤ q + kn − (b − 1)
b
b
kn − (b − 1)
or equivalently,
(b − 1)
µ¹
º j k¶
q + kn
q
−
≤ kn.
b
b
Because of (5.3), it is enough to show that
º j k
¹
³j n k
´
q
q + kn
−
≤k
+1 .
b
b
b
¥n¦
For n = b b + t, 0 ≤ t < b,
jnk
³ jnk jq k
´
³
jq k ´
kn + q = k
b + kt + q = k
+
+ k b + k(t − b) + q −
b .
b
b
b
b
¥ ¦
As, k(t − b) + (q − qb b) < b, then
º
¹
jnk jq k
q + kn
≤k
+
+ k,
b
b
b
completing the proof.
T HEOREM 5.7. Let S = {0, 1, b} ∈ Sn,3 . Then S is 0-periodic if and only if one of the
following conditions is satisfied:
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i) b is ajdivisor
of n;
k
n
ii) b = j + 1, for some nonnegative integer j. In this case, j is unique and is given
by ⌊n/b⌋ + 1.
Proof. If b is a divisor of n, then Lemma 5.6 implies that S is 0-periodic. Now suppose
that b is not a divisor of n. Let i = ⌊n/b⌋ and n = bi + t, with 0 < t < b. If t < i then
b = ⌊ ni ⌋, otherwise b < ⌊ ni ⌋. Since n = (i + 1)b + (t − b) and t − b < 0, it follows that
n
⌊ i+1
⌋ < b. Therefore,
º
¹
jnk
n
<b≤
.
i+1
i
j
k
n
Suppose that b = i+1
+ 1. Since
b−1=
¹
º
n
n
≤
,
i+1
i+1
we have
n ≥ (b − 1)(i + 1).
Therefore, by Lemma 5.6, S is 0-periodic.
j
k
¥ ¦
n
+ 2. Then
Now suppose that ni ≥ b ≥ i+1
n
<b−1
i+1
and, by Lemma 5.6, S is not 0-periodic.
6. The conjecture for bases in Sn,3 . In this section, we prove the following result.
T HEOREM 6.1. Let S ∈ Sn,3 , n ≥ 3. Conjecture 1 holds if S satisfies one of the
following conditions:
i) S is equivalent to {0, a, b}, where a is a divisor
n of n
j and ak ≥ cno;
ii) S is equivalent to {0, 1, b} for some b ≤ min pn , cnn−1 − 1 ;
iii) S is equivalent to a 0-periodic basis.
The rest of this section is dedicated to the proof of Theorem 6.1.
We first observe that, in general, if i and j are positive numbers, then
n
n
+i< +j
i
j
can be written as
(i − j)(n − ij) > 0,
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which is equivalent to
n no
n no
j < min i,
∨ j > max i,
.
i
i
L EMMA 6.2. Let
j Sk = {0, a, b} ∈ Sn,3 , where a is a divisor of n such that a ≥ cn . Then
either order(S) ≤ cnn + cn − 2 or
n
n
− 1 ≤ order(S) ≤ + j − 2,
j
j
(6.1)
for some j ∈ {1, 2, . . . , cn − 1}.
Proof. Note that
j kn 6= 3. If n = 4 then cn = a = 2 and order(S) = 2, which implies
that order(S) ≤ cnn + cn − 2. If n = 8 then cn = 3 and a = 4; a direct computation
considering all possible values of b, namely 1, 3, 5, 7, shows that order(S) = 4 (in fact, in
this case, S ∼ {0, 1, b′ } for some b′ ∈ Zn ).jThen
k (6.1) holds with j = 2. Now suppose that
n 6= 4 and n 6= 8. Suppose that order(S) >
n
cn
+ cn − 2. By Theorem 5.2,
a − 1 ≤ order(S) ≤
n
+ a − 2.
a
Then
n
n
+ cn < + a.
cn
a
Taking into account the observation before this lemma and the fact that for n 6= 3, 4, 8,
cn < n/cn , it follows that a > n/cn , as, by hypothesis, a ≥ cn . Then, because a divides n,
a = ni for some i ∈ {2, 3, . . . , cn − 1}. Then (6.1) holds with j = i.
L EMMA 6.3. Let S = {0, 1, b} ∈ Sn,3 , with b ≤ pn . If S is 0-periodic and b ≥
⌊n/cn ⌋ + 2, then there exists i ∈ [2, cn − 1] such that
i+
jnk
i
− 3 ≤ order(S) ≤ i +
jnk
i
− 2,
Proof. Note that n > 3 and n 6= 8. If b is a divisor of n, then b = n/i for some
i ∈ [2, cn − 1]. By Corollary 5.4, order(S) = ni + i − 2.
If b is not a divisor of n, then, taking into account Theorem 5.7 , b = ⌊n/i⌋ + 1 for some
i ∈ [2, cn − 1]. Let m = ⌊n/i⌋ and n = mi + t, 0 ≤ t < i. By Corollary 5.4,
order(S) =
¹
º j k
¹
º
jnk
t−i
n
mi + t
+
−1=
+i+
− 1.
m+1
i
m+1
i
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655
If m + 1 ≥ i − t, then
¹
º
t−i
= −1.
m+1
¹
º
t−i
= −2,
m+1
If m + 1 < i − t, then
j k
j k
as i − t ≤ i ≤ cn − 1 ≤ cnn − 1 ≤ b − 3 ≤ 2b = 2m + 2. Note that cn ≤ cnn for
n 6= 3, 8. Thus, the result follows.
h
j k
i
L EMMA 6.4. Let S = {0, 1, b} ∈ Sn,3 . If b ∈ cn + 1, cnn + 1 , then order(S) ≤
j k
n
cn + cn − 2.
Proof. By Theorem 5.3, order(S) ≤
¥n¦
+ b − 2. Thus, it is enough to show that
¹ º
jnk
n
.
(6.2)
≤ cn +
b+
b
cn
b
Taking into account the observation before Lemma 6.2, if cn < min{b, nb } then
jnk
n
n
≤ b + < cn + ,
b+
b
b
cn
which implies (6.2), as cn is an integer. Since
j k
(
n no
cn < b ≤ cnn
cn < min b,
⇔
b
cn < b ≤ n − 1
if cn is not a divisor of n
,
if
cn is a divisor of n
cn
j k
we now need to show that (6.2) holds if either b = cnn + 1 or b = cnn and cn divides n. The
latter is immediate. For the first case, note that






n
 = cn − 1,
j k
n
+
1
cn
as, if n =
j
n
cn
k
cn + t, with 0 ≤ t < cn , then
µ¹
º
¶ µ
¹ º
¶
n
n
+ 1 + t − cn +
+1 ,
cn
cn
j k
j k
with 0 ≤ t − cn + cnn + 1 < cnn + 1.
n = (cn − 1)
Next we give a result that allows us to show that the conjecture holds if S = {0, 1, b},
with b ≤ pn and b ∈ [⌊n/cn ⌋ + 2, ⌊n/(cn − 1)⌋ − 1]. Note that ⌊n/cn ⌋ + 2 ≤ pn if and only
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if cn ≥ 3. Also, by Lemma 2.5, for cn ≥ 3 the previous interval is nonempty if and only if
n = 14 or n ≥ 16. Finally, observe that ⌊n/b⌋ = cn − 1.
We think the method used to prove the conjecture in this case might be generalizable to
n
the cases in which b ∈ [⌊ cnn−k ⌋, ⌊ cn −k+1
⌋ − 1], with 1 ≤ k ≤ cn − 3, when this interval is
nonempty. Some results presented in Section 2 will be used.
h
i
L EMMA 6.5. Let n be a positive integer such that cn ≥ 3, b ∈ ⌊ cnn ⌋ + 2, ⌊ cnn−1 ⌋ − 1
and t = n − (cn − 1)b. If
¹
º
n
+ 1 < b − t,
cn
(6.3)
√
√
then either cn = ⌊ 3 n⌋, or cn = ⌊ 3 n⌋ + 1 and n = c3n − 1. Moreover, 3cn ≤ ⌊n/cn ⌋ + 2.
jProof. kLet m = cn − 1 and r =j ⌊n/cnk⌋ + cn − 2. First we show that if (6.3) holds, then
b = cnn−1 − 1. Suppose that b ≤ cnn−1 − 2. Then t ≥ 2(cn − 1) and, taking into account
Lemma 2.6, we get
¹
º
¹ º
n
n
b−t≤
− 2 − 2(cn − 1) ≤
+ 1,
cn − 1
cn
j
k
a contradiction. Now suppose that b = cnn−1 − 1 and (6.3) holds. Then t ≥ cn − 1. If
√
cn = ⌊ 3 n⌋ + 1 and n ≤ c3n − 2, then, taking into account Lemma 2.6,
¹
º
¹ º
n
n
b−t≤
− 1 − (cn − 1) ≤
+ 1,
cn − 1
cn
√
√
a contradiction. Thus, cn = ⌊ 3 n⌋ or cn = ⌊ 3 n⌋ + 1 and n = c3n − 1. Taking into account
Lemma 2.7, the result follows.
L EMMA 6.6. Let S = {0, 1, b} ∈ Sn,3 , with cn ≥ 3. Suppose that b ∈ [⌊n/cn ⌋ +
2, ⌊n/(cn − 1)⌋ − 1]. Then
¹ º
n
+ cn − 2.
order({0, 1, b}) ≤
cn
Proof. Note that nj = k14 or n ≥ 16 for the interval [⌊n/cn ⌋ + 2, ⌊n/(cn − 1)⌋ − 1] not
to be empty. Let r = cnn + cn − 2 and n = (cn − 1)b + t for some 0 < t < b. Note that
t ≥ cn − 1. Let m = ⌊n/b⌋ = cn − 1. Since cn ≤ ⌊n/cn ⌋, 2m ≤ r.
Let
k1 (i) = ib and k2 (i) = i(b − 1) + r, for i ∈ {0, 1, . . . , m},
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k1 (i) = ib − n and k2 (i) = i(b − 1) + r − n, for i ∈ {m + 1, m + 2, . . . , 2m},
and
k1 (i) = ib − 2n and k2 (i) = i(b − 1) + r − 2n, for i ∈ {2m + 1, 2m + 2, . . . , 3m}.
Consider the following intervals in Z : Ii = [k1 (i), k2 (i)], i ∈ {0, 1, . . . , 3m}. Note that, for
each i = 0, 1, . . . , 3m, k1 (i) < k2 (i) and k1 (i) < n. Also, 0 ≤ k1 (i), for i = 0, 1, . . . , 3m,
Sr
i 6= 2m + 1. We have rS ≡ i=0 Ii (mod n). We next show that if
then
S2m
i=0 Ii
b − t ≤ ⌊n/cn ⌋ + 1,
(6.4)
⌊n/cn ⌋ + 1 < b − t,
(6.5)
≡ Zn (mod n); if
S3m
then i=0 Ii ≡ Zn (mod n), which implies rS = Zn . Note that 2m < r and, by Lemma 6.5,
if (6.5) holds, 3m ≤ r.
Consider the intervals Ii , i = 0, 1, . . . , 3m, ordered in the following way:
I0 , I2m+1 , Im+1 , I1 , I2m+2 , Im+2 , . . . , I3m , I2m , Im .
Clearly, for j = 1, 2, . . . , m,
k1 (j − 1) ≤ k1 (m + j) ≤ k1 (j) and k1 (2m + j) ≤ k1 (m + j).
We show that, for each j = 1, 2, . . . , m,
(i)
(ii)
(iii)
(iv)
(v)
k2 (m + j) + 1 ≥ k1 (j);
k2 (j − 1) + 1 ≥ k1 (m + j) if (6.4) holds;
k2 (j − 1) + 1 ≥ k1 (2m + j) ≥ k1 (j − 1) if (6.5) holds;
k2 (2m + j) + 1 ≥ k1 (m + j) if (6.5) holds;
k2 (m) ≥ n − 1,
which completes the proof.
Condition (i) follows easily taking into account that cn + t ≤ ⌊ cnn ⌋ + 1, as
t − (cn − 1)b ≤ n − (cn − 1)
µ¹
¹ º¶ ¹ º
º
¶ µ
n
n
n
+ 2 = n − cn
+
− 2(cn − 1)
cn
cn
cn
º
¹ º
n
n
− 2(cn − 1) =
− cn + 1.
≤ cn − 1 +
cn
cn
¹
Condition (ii) follows from a simple calculation.
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Now suppose that (6.5) holds. The first inequality in condition (iii) holds as
º
¹ º
¹ º
n
n
n
− 1 − 2(cn − 1) ≤
+1≤
+ cn − j,
b − 2t ≤
cn − 1
cn
cn
¹
where
k the second inequality follows from Lemma 2.6. Since we have shown in (i) that t ≤
j
n
cn − cn + 1, then
b > ⌊n/cn ⌋ + 1 + t > 2t,
which implies the second inequality.
√
Condition (iv) holds if 2cn + t ≤ ⌊n/cn ⌋ + 2. By Lemma 6.5 either cn = ⌊ 3 n⌋ or
√
√
cn = ⌊ 3 n⌋ + 1 and n = c3n − 1. If cn = ⌊ 3 n⌋, by Lemma 2.6,
t≤
¹
º
¹ º
n
n
−1−
− 1 ≤ cn + 3 − 2.
cn − 1
cn
Thus, taking into account Lemma 2.7,
¹
º
n
2cn + t ≤ 3cn + 1 ≤
+ 2.
cn
√
If cn = ⌊ 3 n⌋ + 1 and n = c3n − 1, by Lemma 2.6,
t≤
¹
º
¹ º
n
n
−1−
− 1 ≤ cn + 2 − 2.
cn − 1
cn
By Lemma 2.7,
¹
º
n
2cn + t ≤ 3cn ≤
+ 2.
cn
Finally, note that condition (v) is equivalent to t ≤ ⌊n/cn ⌋, which holds as cn ≥ 3 and
we have shown that t + cn ≤ ⌊n/cn ⌋ + 1.
Proof of Theorem 6.1. It follows from Lemma 6.2 that if condition i) holds then Conjecture 1 is satisfied.
Now suppose that S = {0,j1, b},
k with b ≤ pn . If b ≤ cn , Conjecture 1 holds by
n
Theorem 5.3; if b ∈ [cn + 1, cn + 1], the conjecture holds by Lemma 6.4; if b ∈
hj k
k
i
j
n
n
cn + 2, cn −1 − 1 then cn ≥ 3 and Conjecture 1 holds by Lemma 6.6. Finally, if
b ≥ ⌊n/cn ⌋ + 2 and S is 0 -periodic, Conjecture 1 holds by Lemma 6.3. Since two equivalent
bases have the same order, the result follows.
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659
7. The conjecture for bases with cardinality larger than 3. In this section, we include
some partial results regarding Conjecture 1 for bases with cardinality larger than 3.
The next lemma shows that to prove Conjecture 1 it is enough to consider bases S for
Zn such that
|S| ≤ max
½
n
d
µ¹
º
¶
¾
d−2
+ 1 : d|n, d ≥ ⌊n/cn ⌋ + cn − 1 .
⌊n/cn ⌋ + cn − 3
L EMMA 7.1. [7] Let n be a positive integer and r ∈ [2, n − 1]. Let S ∈ Sn be such that
order(S) ≥ r. Then
|S| ≤ max
½
n
d
µ¹
º
¶
¾
d−2
+ 1 : d|n, d ≥ r + 1 .
r−1
C OROLLARY 7.2. If S ∈ Sn is equivalent to {0, 1, s1 , . . . , sr } ∈ Sn , with 1 < s1 <
· · · < sr , then
¹
n
sr
º
≤ order(S) ≤
min
i∈{1,2,...,r}
½¹
º
¾
n
+ si − 2 .
si
Proof. The proof of the inequality on the left is analogous to the one given in the proof
of the left inequality in Theorem 5.3. The inequality on the right follows from the fact that
order(S) ≤ mini {order({0, 1, si })} and Theorem 5.3.
We then have the following consequence of Corollary 7.2, Lemmas 6.4 and 6.6 and
Theorem 5.3.
C OROLLARY 7.3. If Sh ∈ S
to {0, 1, s1 , . . . , sr } and there exists i ∈
j n is equivalent
k
i
n
{1, 2, . . . , r} such that si ∈ cn , cn −1 − 1 , then S satisfies Conjecture 1.
C OROLLARY 7.4. If S ∈ Sn is equivalent to S ′ = {0, s1 , . . . , sr } and S ′ contains an
element of order m, with m ∈ {cn , . . . , cnn }, then S satisfies Conjecture 1.
Proof. By Lemma 4.4,
order(S) ≤
n
+ m − 2.
m
Taking into account the observation before Lemma 6.2, if m is a divisor
j k of n such that m ∈
n
n
n
n
{c, . . . , c }, then m + m ≤ cn + cn , which implies that m + m ≤ cnn + cn and the result
follows.
Electronic Journal of Linear Algebra ISSN 1081-3810
A publication of the International Linear Algebra Society
Volume 20, pp. 640-660, October 2010
ELA
http://math.technion.ac.il/iic/ela
660
M.I. Bueno and S. Furtado
8. Conclusions and open problems. Given a positive integer n ≥ 3, we defined the
critical point cn and conjectured that all intervals of the form [⌊n/i⌋ + i − 1, ⌊n/(i − 1)⌋ − 2],
with 2 ≤ i < cn , are gaps in the set En of orders of bases for Zn . It was already known that
bases with cardinality 2 have order n − 1, and therefore, they satisfy our conjecture.
In this paper, we have proven some partial results regarding bases of cardinality 3 and
larger. The main result is Theorem 23. However, there are many open questions still to
answer. For bases with cardinality 3, it needs to be proven that the conjecture holds when a
basis S is equivalent to {0, 1, b} with b ∈ [⌊n/(cn − 1)⌋, pn ], where pn is defined in (5.1).
We think that in order to prove this result the concept of K-periodicity needs to be studied in
greater detail. If a basis S is equivalent to {0, a, b}, where a is a divisor of n, a 6= 1, it is
still an open question if the conjecture holds when a < cn . Though we did not show that all
bases for Zn with cardinality 3 are equivalent to a set of the form {0, 1, b} or {0, a, b}, where
a 6= 1 is a divisor of n, at least for almost all bases this seems to happen. If there exist bases
for Zn which are not equivalent to sets of any of those two types, the conjecture should also
be proven for them.
Finally, we have only proven that the conjecture holds for very specific bases for Zn with
cardinality larger than 3, so there is a lot to be done concerning those bases.
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